RE: [PHP] ImageColorAllocate() Problem

I think it might be because you're passing a string to the function instead of a hex value... try changing it to this and see if it works $color = ImageColorAllocate($im, hexdec($hex1), hexdec($hex2), hexdec($hex3)); HTH Martin -----Original Message----- From: Jed R. Brubaker [mailto:jed.brubaker@psych.utah.edu] Sent: Friday, 3 October 2003 4:23 PM To: php-general@lists.php.net Subject: [PHP] ImageColorAllocate() Problem I am having a bizzare problem with the imagecolorallocate() function. Maybe all of you can help me! Here is the code: $hex1 = "0x".$color{0}.$color{1}; $hex2 = "0x".$color{2}.$color{3}; $hex3 = "0x".$color{4}.$color{5}; $color = ImageColorAllocate($im, $hex1, $hex2, $hex3); Simple enough! $color starts off as a hex color code minus the #, and then is pulled into each of the 3 alpha values by way for the hex variables. It is later outputted in text: ImageTTFText ($im, $cursive_size, 0, $text1_x, $text1_y, $color, $font1, $cursive_text); Problem: The text is always black. Something strange: If I just type in a randomly selected color, it works! Is there some reason that the values in my variables wouldn't be passing? Thanks in advance, Jed R. Brubaker -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php __________ Information from NOD32 1.518 (20030925) __________ This message was checked by NOD32 for Exchange e-mail monitor. http://www.nod32.com